In germanium the energy gap is about 0.75 eV. The wavelength of light which germanium starts absorbing is

Options

A. 5000 Å
B. 1650Å
C. 16500Å
D. 165000Å

Explanation

To find the wavelength of light which germanium starts absorbing, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Energy Gap**:
The energy gap (E) for germanium is given as 0.75 eV. This energy gap is the minimum energy required to excite an electron from the valence band to the conduction band.

2. **Using the Energy-Wavelength Relationship**:
The energy of a photon can be related to its wavelength using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy in joules,
- \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

3. **Converting Energy from eV to Joules**:
Since the energy is given in electron volts, we need to convert it to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore, the energy in joules is:
\[ E = 0.75 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.2 \times 10^{-19} \, \text{J} \]

4. **Rearranging the Formula to Solve for Wavelength**:
Rearranging the energy-wavelength relationship gives us:
\[ \lambda = \frac{hc}{E} \]

5. **Substituting the Values**:
Now we can substitute the values of \(h\), \(c\), and \(E\):
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{1.2 \times 10^{-19} \, \text{J}} \]

6. **Calculating the Wavelength**:
Performing the calculation:
\[ \lambda = \frac{1.989 \times 10^{-25}}{1.2 \times 10^{-19}} \approx 1.6575 \times 10^{-6} \, \text{m} \]
Converting meters to angstroms (1 m = \(10^{10}\) angstroms):
\[ \lambda \approx 1.6575 \times 10^{-6} \, \text{m} \times 10^{10} \, \text{angstrom/m} \approx 16575 \, \text{angstrom} \]

7. **Final Answer**:
Therefore, the wavelength of light which germanium starts absorbing is approximately **16500 angstroms**.